# How do you differentiate the following parametric equation:  (sint,cost-t)?

Nov 16, 2016

$\frac{d}{\mathrm{dt}} \left(\sin t , \cos t - t\right) = - \tan t - \csc t$

#### Explanation:

Define $f \left(t\right) , x \left(t\right) , y \left(t\right)$ as follows

$f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ where $\left\{\begin{matrix}x \left(t\right) = \sin t & \implies \frac{\mathrm{dx}}{\mathrm{dt}} = \cos t \\ y \left(t\right) = \cos t - t & \implies \frac{\mathrm{dy}}{\mathrm{dt}} = - \sin t - 1\end{matrix}\right.$

Then by the chain rule we have;

$f ' \left(t\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin t - 1}{\cos t}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \sin \frac{t}{\cos} t - \frac{1}{\cos} t$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \tan t - \csc t$